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Homework Help: Calculate Refraction without Snell's Law?

  1. Mar 20, 2009 #1
    Ok, I'm decided to post this here just because after searching for refraction of light, I see alot of posts in this category. Currently in my Physics class, we're working on the theory of light in the book, Newton to Einstein the trail of light. After covering refraction and snell's law to calculate refraction of a crest of light when it passes from one substance to another, I found another way to calculate the refraction without Snell's law. Please comment on what I have here as I can't seem to locate any information regarding it.

    1. The problem statement, all variables and given/known data

    Hopefully my attachment will work, it's a Word file I just drew. The first sketch shows a light wave at one instance in time heading toward a certain transparent solid. The speed of light in the solid is 2 x 10^10 cm/second. In a vacuum it's 3 x 10^10 cm/second. We were supposed to reproduce the sketch and add the crest line as it would appear 10^-10 seconds later, ignoring the reflected line.

    2. Relevant equations

    I roughly drew out Snell's law where one is supposed to find the angle of incidence (i in the drawing) the a perpendicular line towards the solid one wavelength long. where it touches the boundry line, the refracted crestline would be drawn at the angle of refraction as calculated by Snell's law using the Index of refraction of the two substances and so on which you can draw a perpendicular line from this line to the boundary line which would measure one wavelength of the light in the solid.

    3. The attempt at a solution
    My thoughts were, since with the given information in the problem, we don't know the angle of incidence or the index of refraction for the substances (although they could easily be calculated) why not use the crest line given and draw two perpendicular lines from it into the solid, one from point B would be 3cm long since it would travel that far in the given time in air and from point A 2cm long since in the solid the light would travel that distance. The the new crest in air would remain parallel to the old crest and the point at the boundary line would then be used to draw the crest line in the solid along with the point 2cm in.

    I have drawn this up many different times using Snell's law and different angles checking it with merely drawing the distance lines as I've shown and the results are very nearly accurate if not accurate, Exact measurements with a cheap protrator and ruler are hard to acheive as well as duplicate. Just my thinkings on this, please comment with what you think. In class, going over this problem, my instructor actually began solving the problem just as I described until another student said snell's law must be used, according to the book. After looking at the problem for a minute, he changed his method to use Snell's law. The next day before class, I showed him the sketches I'd made, telling him I agreed with how he initially began the problem and felt it a viable route. He seemed to stick with Snell's law stating that he had mispoke before.

    Attached Files:

  2. jcsd
  3. Mar 20, 2009 #2


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  4. Mar 20, 2009 #3
    Those drawings look accurate to me. Perhaps you should try using them to derive Snell's law?
  5. Mar 20, 2009 #4


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    Hi deadendcustom,

    That's a good idea to consider, but I don't believe this idea is quite right. The problem is that the distance between the wave crests needs to be measured in the direction of the travel of the wave. If you look at this picture:

    http://img5.imageshack.us/img5/2346/snells.jpg [Broken]

    on the left is how the "standard" snell's law problem would be set up, and the right shows how this is related to the wavecrests. The important thing is that the distance lambda2 is measured perpendicular to the wavecrests, which is not occuring in your diagram. (Remember that this measured direction is in the same direction as the refracted wave, but you have it measured in the direction of the incident wave.)

    But I think there is a good reason why it appears to work for many cases. In the diagram below, I have drawn a circle with radius 2cm, and I have shown 2 different cases. For many indices of refraction and angles of incidence, you would get something like in the left diagram:

    http://img5.imageshack.us/img5/8185/snells2.jpg [Broken]

    The 2cm distance from your diagram has to be a radius of this circle. (Your 2cm line is a continuation of the initial ray, while the "real" one is given by Snell's law.) But as the diagram on the right shows, you can also come up with cases where it would give a very different result.

    I think this is a fun thing to think about though. You could derive a Snell-law type relation for things that followed your idea, and even find some expressions for how accurate it is for different ranges of n and theta. It's definitely a neat and straightforward way to draw the diagram.

    (Sorry about the large diagrams!)
    Last edited by a moderator: May 4, 2017
  6. Mar 20, 2009 #5

    Exactly what I was looking for, thank you for your time. I don't have a decent drafting program so I was drawing my sketches by hand. I tested the way I proposed starting with 4 different angles of incidence and came up with crests very close to each other as your first drawing shows. Without accurate measuring equipment I wasn't able to determine if the the two crests were actually different or if my measuring tools were creating the differences. I assumed they were not equal but at least very near to each other which would make for a quick and simple way of estimating the refraction. I hadn't used as large of an angle of incidence as your second drawing yet so I hadn't noticed the increasing error factor.

    When my internet search turned up no other results for a method other than Snell's law, I figured there would be a flaw in there somehow. I'm happy to finally have an answer on the subject though.

  7. Mar 20, 2009 #6


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    I just want to say that even when I was writing my response, I was hoping that I wasn't in any way being discouraging. Of course I have no idea what your longterm plans are, but just thinking about this idea tells me that you have your brain turned on and engaging the physics. Of course nobody like being wrong, but thinking about these kinds of ideas is a fantastic way to really understand what's really going on behind the formulas.

    All the scientists I know are always trying to think of new things, and most of it turns out to not be useful, but just disproving it helps you understand things more. I enjoyed reading your idea and thinking about it, and encourage you to keep thinking beyond your course.
  8. Mar 20, 2009 #7
    I can easily see where some could get offended or upset when they are proven wrong, but as you stated, one can learn a great deal simply by being wrong. I'm actually finally in my first semester of college after graduating high school 9 years ago. Taking an intro to Physics class working towards a degree in Engineering so you can bet I'll definitely be spending alot of time on this forum.
  9. Mar 20, 2009 #8


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    That sounds great, and welcome to PF!
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